integral of (5x-5)/(x^2-x-6)

解

\int \frac{5 x - 5}{x ^{2} - x - 6} d x

\frac{5}{2} ln 4 x^{2} - 4 x - 24 - \frac{1}{2} (ln \frac{2}{5} x + \frac{4}{5} - ln \frac{2}{5} x - \frac{6}{5}) + ln ∣ x + 2 ∣ - ln ∣ x - 3 ∣ + C

解答ステップ

\int \frac{5 x - 5}{x ^{2} - x - 6} d x

拡張

\frac{5 x - 5}{x ^{2} - x - 6} : \frac{5 x}{x ^{2} - x - 6} - \frac{5}{x ^{2} - x - 6}

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{5 x}{x ^{2} - x - 6} d x - \int \frac{5}{x ^{2} - x - 6} d x

\int \frac{5 x}{x ^{2} - x - 6} d x = 10 (\frac{1}{4} ln 4 x^{2} - 4 x - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} x - ln - \frac{6}{5} + \frac{2}{5} x))

\int \frac{5}{x ^{2} - x - 6} d x = - ln ∣ x + 2 ∣ + ln ∣ x - 3 ∣

= 10 (\frac{1}{4} ln 4 x^{2} - 4 x - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} x - ln - \frac{6}{5} + \frac{2}{5} x)) - (- ln ∣ x + 2 ∣ + ln ∣ x - 3 ∣)

簡素化

10 (\frac{1}{4} ln 4 x^{2} - 4 x - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} x - ln - \frac{6}{5} + \frac{2}{5} x)) - (- ln ∣ x + 2 ∣ + ln ∣ x - 3 ∣) : \frac{5}{2} ln 4 x^{2} - 4 x - 24 - \frac{1}{2} (ln \frac{2}{5} x + \frac{4}{5} - ln \frac{2}{5} x - \frac{6}{5}) + ln ∣ x + 2 ∣ - ln ∣ x - 3 ∣

= \frac{5}{2} ln 4 x^{2} - 4 x - 24 - \frac{1}{2} (ln \frac{2}{5} x + \frac{4}{5} - ln \frac{2}{5} x - \frac{6}{5}) + ln ∣ x + 2 ∣ - ln ∣ x - 3 ∣

定数を解答に追加する

= \frac{5}{2} ln 4 x^{2} - 4 x - 24 - \frac{1}{2} (ln \frac{2}{5} x + \frac{4}{5} - ln \frac{2}{5} x - \frac{6}{5}) + ln ∣ x + 2 ∣ - ln ∣ x - 3 ∣ + C